Find equation of motion using. Lagrangian given equation is $L^{\prime }=\frac{m}{2}(a{\dot{x}}^2+2b\dot{x}y+c{\dot{y}}^2)-\frac{k}{2}(a{x}^2+2bxy+c{y}^2)$

$$L^{\prime }=\frac{m}{2}(a{\dot{x}}^2+2b\dot{x}y+c{\dot{y}}^2)-\frac{k}{2}(a{x}^2+2bxy+c{y}^2)$$
where a, b, and c are arbitrary constants but subject to the condition that $b^2-ac\ne 0$. What are the equations of motion? Examine particularly the two cases $a=0=c$ and $b=0$, $c=-a$. What is the physical system described by the above Lagrangian? Show that the usual Lagrangian for this system as defined by Eq. (1.56) is related to $L^{\prime }$ by a point transformation (cf. Derivation 10). What is the significance of the condition on the value of $b^2-ac$?

$L=T-U\text{Eq.(1.56)}$

It's two dimensional. If there was only $x$ than, it was easier to differentiate. But, I was thinking what I should do now. Then, all the things I had done that is differentiating respect to $x$ and $y$ both.

$$\frac{d^{2}L}{dxdy}=-kax-bkx-bky-cky$$ $$\frac{d^{2}L}{d\dot{x}d\dot{y}}=am\dot{x}+bm\dot{y}+bm\dot{x}+cm\dot{y}$$ $$\frac{d}{dt}(\frac{d^{2}L}{d\dot{x}d\dot{y}})=am\ddot{x}+bm\ddot{y}+bm\ddot{x}+cm\ddot{y}$$ $$\frac{d}{dt}(\frac{d^{2}L}{d\dot{x}d\dot{y}})-\frac{d^2}{dxdy}=0$$ $$am\ddot{x}+bm\ddot{y}+bm\ddot{x}+cm\ddot{y}=-(kax+bkx+bky+cky)$$ How can I find equation of motion for that? (What are the equations of motion?) Should I find a equation for $x$ at first then another equation for $y$?